Problem 55
Question
find and simplify the difference quotient $$ \frac{f(x+h)-f(x)}{h}, h \neq 0 $$ for the given function. $$ f(x)=4 x $$
Step-by-Step Solution
Verified Answer
The simplified difference quotient is \(4\).
1Step 1: Insert Function into Difference Quotient Formula
Replace \(f(x)\) and \(f(x+h)\) in the difference quotient formula \(\frac{f(x+h)-f(x)}{h}\) with the given function. The function given is \(f(x)=4x\), so substituting these into the formula gives \(\frac{4(x+h)-4x}{h}\).
2Step 2: Simplifying the Numerator
Next, the expression in the numerator, \(4(x+h)-4x\), should be simplified. Simplifying this gives \(4x+4h-4x\). When the like terms are combined, we get the simplified form of \(4h\).
3Step 3: Simplify the Quotient
Returning the simplified numerator back into the quotient gives us \(\frac{4h}{h}\). The \(h\) in the numerator and denominator will cancel each other out, leaving 4 as the final simplified difference quotient.
Key Concepts
Algebraic SimplificationFunction SubstitutionNumerical Simplification
Algebraic Simplification
Algebraic simplification involves reducing expressions to their simplest form by performing basic algebraic operations like addition, subtraction, multiplication, and division, as well as combining like terms and canceling common factors. This process is particularly helpful when working with the difference quotient, which can often appear more complex than it actually is.
For example, in the given function in our exercise, we start by inserting our function into our difference quotient, which leads to an expression that might look complicated at first. But by systematically applying algebraic simplification, as seen in Step 2 of our solution, we combine like terms, in this case, the terms involving 'x'. When the like terms, +4x and -4x, are combined, they effectively cancel each other out, leaving us with a much simpler expression of 4h.
By conducting these steps methodically and combining terms wisely, algebraic simplification not only makes the difference quotient more manageable but also prepares the expression for further simplification, such as canceling out common factors, which we see in the following steps.
For example, in the given function in our exercise, we start by inserting our function into our difference quotient, which leads to an expression that might look complicated at first. But by systematically applying algebraic simplification, as seen in Step 2 of our solution, we combine like terms, in this case, the terms involving 'x'. When the like terms, +4x and -4x, are combined, they effectively cancel each other out, leaving us with a much simpler expression of 4h.
By conducting these steps methodically and combining terms wisely, algebraic simplification not only makes the difference quotient more manageable but also prepares the expression for further simplification, such as canceling out common factors, which we see in the following steps.
Function Substitution
Function substitution is an essential concept for calculating the difference quotient. It requires replacing the variable 'x' in a function with another expression. This concept is key to understanding and working with functions algebraically, and it's used to find the rate of change of our function at a particular point.
Looking at Step 1 of our solution, we substitute 'x' and 'x+h' into our original function, which is described by f(x)=4x. In the context of the difference quotient \(\frac{f(x+h)-f(x)}{h}\), we effectively replace 'x' with 'x+h' within our function to find f(x+h), which yields 4(x+h). Then, we perform the same substitution using only 'x' to find f(x), which is simply 4x. Both of these substituted expressions are used in the difference quotient formula, setting the stage for the algebraic simplification that follows.
Substitution is a pivotal technique used across many areas of mathematics, and mastering it can help clarify the process of manipulating functions and expressions for various applications.
Looking at Step 1 of our solution, we substitute 'x' and 'x+h' into our original function, which is described by f(x)=4x. In the context of the difference quotient \(\frac{f(x+h)-f(x)}{h}\), we effectively replace 'x' with 'x+h' within our function to find f(x+h), which yields 4(x+h). Then, we perform the same substitution using only 'x' to find f(x), which is simply 4x. Both of these substituted expressions are used in the difference quotient formula, setting the stage for the algebraic simplification that follows.
Substitution is a pivotal technique used across many areas of mathematics, and mastering it can help clarify the process of manipulating functions and expressions for various applications.
Numerical Simplification
Numerical simplification is the final stage of our difference quotient problem that involves simplifying the mathematical expressions to their numerical representations. After algebraic simplification has been performed and all like terms have been combined, what remains in the expression may often still contain variables. The aim is now to further simplify the expression to find the most reduced numerical form possible.
In Step 3 of our solution, after we have simplified our numerator to 4h through algebraic manipulation, we reintroduce it into the difference quotient where we encounter a fraction where 'h' is both in the numerator and the denominator. This presents an opportunity for numerical simplification, as we can cancel the common factors. Because 'h' ≠ 0, we are allowed to divide both the numerator and denominator by 'h', which leaves us with the number 4.
Numerical simplification brings us to the simplest form of the expression; in our exercise, it helps to reveal the constant rate at which our linear function f(x)=4x changes, regardless of the value of 'x'. Understanding this part of the process is crucial to interpreting the results and implications of problems involving rates of change.
In Step 3 of our solution, after we have simplified our numerator to 4h through algebraic manipulation, we reintroduce it into the difference quotient where we encounter a fraction where 'h' is both in the numerator and the denominator. This presents an opportunity for numerical simplification, as we can cancel the common factors. Because 'h' ≠ 0, we are allowed to divide both the numerator and denominator by 'h', which leaves us with the number 4.
Numerical simplification brings us to the simplest form of the expression; in our exercise, it helps to reveal the constant rate at which our linear function f(x)=4x changes, regardless of the value of 'x'. Understanding this part of the process is crucial to interpreting the results and implications of problems involving rates of change.
Other exercises in this chapter
Problem 54
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complete the square and write the equation in standard form. Then give the center and radius of each circle and graph the equation. $$ x^{2}+y^{2}-10 x-6 y-30=0
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Find a. \((f \circ g)(x) \quad \) b. \((g \circ f)(x) \quad \) c. \((f \circ g)(2) \quad \) d. \((g \circ f)(2)\) $$f(x)=4 x-3, g(x)=5 x^{2}-2$$
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Begin by graphing the standard quadratic function, \(f(x)=x^{2} .\) Then use transformations of this graph to graph the given function. $$g(x)=(x-2)^{2}$$
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